Question

question 10
evaluate the limit: \\(\lim\limits_{x\to 5} \dfrac{x - 3}{\sqrt{2x - 1}}\\)
\\(\bigcirc\\) no correct answer choice is given.
\\(\bigcirc\\) \\(2x - 7\\)
\\(\bigcirc\\) \\(\dfrac{6}{\sqrt{11}}\\)
\\(\bigcirc\\) 4
\\(\bigcirc\\) 2/3
\\(\bigcirc\\) does not exist

Explanation:

Step1: Identify the limit type

This is a limit of a rational function (a quotient of two functions) as \( x \to 5 \). We can use direct substitution if the denominator is not zero at \( x = 5 \).

Step2: Substitute \( x = 5 \) into the numerator and denominator

• Numerator: \( x - 3 \) at \( x = 5 \) is \( 5 - 3 = 2 \).
• Denominator: \( \sqrt{2x - 1} \) at \( x = 5 \) is \( \sqrt{2(5) - 1} = \sqrt{10 - 1} = \sqrt{9} = 3 \). Wait, wait, no, wait: \( 2x - 1 \) when \( x = 5 \) is \( 10 - 1 = 9 \), so \( \sqrt{9} = 3 \). Wait, but let's recalculate:

Wait, numerator: \( 5 - 3 = 2 \)

Denominator: \( \sqrt{2(5) - 1} = \sqrt{10 - 1} = \sqrt{9} = 3 \)

So the limit is \( \frac{2}{3} \)? Wait, but let's check again. Wait, the original function is \( \frac{x - 3}{\sqrt{2x - 1}} \). So substituting \( x = 5 \):

Numerator: \( 5 - 3 = 2 \)

Denominator: \( \sqrt{2(5) - 1} = \sqrt{10 - 1} = \sqrt{9} = 3 \)

So \( \frac{2}{3} \). Let's check the options. One of the options is \( \frac{2}{3} \). Wait, but wait, did I make a mistake? Wait, let's re-express:

Wait, \( \lim_{x \to 5} \frac{x - 3}{\sqrt{2x - 1}} \). Since the function is continuous at \( x = 5 \) (the denominator is \( \sqrt{2(5) - 1} = \sqrt{9} = 3
eq 0 \)), we can directly substitute \( x = 5 \):

\( \frac{5 - 3}{\sqrt{2(5) - 1}} = \frac{2}{\sqrt{9}} = \frac{2}{3} \). Yes, that's correct.

Answer:

\( \frac{2}{3} \) (the option with \( 2/3 \))